Problem: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $27$ years; the standard deviation is $2.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living between $24.7$ and $33.9$ years.
Answer: $27$ $24.7$ $29.3$ $22.4$ $31.6$ $20.1$ $33.9$ $99.7\%$ $68\%$ $15.85\%$ $15.85\%$ We know the lifespans are normally distributed with an average lifespan of $27$ years. We know the standard deviation is $2.3$ years, so one standard deviation below the mean is $24.7$ years and one standard deviation above the mean is $29.3$ years. Two standard deviations below the mean is $22.4$ years and two standard deviations above the mean is $31.6$ years. Three standard deviations below the mean is $20.1$ years and three standard deviations above the mean is $33.9$ years. We are interested in the probability of a zebra living between $24.7$ and $33.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the zebras will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $68\%$ of the zebras will have lifespans within 1 standard deviation of the mean. The probability of a particular zebra living between $24.7$ and $33.9$ years is ${68\%} + \color{orange}{15.85\%}$, or $83.85\%$.